Question

Sin^2x=3cos^2x+sin2x

Answer 1

Ответ:

x₁=acrctg1/3+πn, n∈Z

x₂=3π/4+πk, k∈Z

Пошаговое объяснение:

sin²x=3cos²x+sin(2x)

Воспользуемся формулой sin 2x:

sin²x=3cos²x+2sin(x)*cos(x)

3cos²x+2sin(x)*cos(x)-sin²x=0

Разделим обе части уравнения на sin²x:

3ctg²x+2ctgx-1=0

Замена ctgx=t:

3t²+2t-1=0

D=4+4*3=16

t₁=(-2+4)/6=1/3

t₂=(-2-4)/6=-1

Обратная замена:

ctgx=1/3          ctgx=-1

Находим корни:

x₁=acrctg1/3+πn, n∈Z

x₂=3π/4+πk, k∈Z

Answer 2

To solve the equation sin^2(x) = 3cos^2(x) + sin(2x), we can use trigonometric identities and algebraic manipulations. Let's go step-by-step:

Step 1: Use the double angle identity for sine: sin(2x) = 2sin(x)cos(x)

The equation becomes: sin^2(x) = 3cos^2(x) + 2sin(x)cos(x)

Step 2: Use the Pythagorean identity: sin^2(x) + cos^2(x) = 1

Rearrange the equation: sin^2(x) = 1 - cos^2(x)

Step 3: Substitute the rearranged equation into the original equation:

1 - cos^2(x) = 3cos^2(x) + 2sin(x)cos(x)

Step 4: Move all the terms to one side of the equation:

1 - 3cos^2(x) - cos^2(x) - 2sin(x)cos(x) = 0

Step 5: Combine like terms:

1 - 4cos^2(x) - 2sin(x)cos(x) = 0

Step 6: Factor out common terms:

1 - 2cos(x)(2cos(x) + sin(x)) = 0

Now we have a factored equation. To solve for x, set each factor equal to zero and solve for cos(x):

1. 2cos(x) + sin(x) = 0

Since we cannot directly solve for cos(x) in this equation, we'll use the Pythagorean identity again:

sin^2(x) + cos^2(x) = 1

Square the equation: sin^2(x) = 1 - cos^2(x)

Substitute this into the first equation: 2cos(x) - (1 - cos^2(x)) = 0

Simplify: 2cos(x) - 1 + cos^2(x) = 0

Rearrange and solve for cos(x): cos^2(x) + 2cos(x) - 1 = 0

Now, we have a quadratic equation in cos(x):

cos(x) = [-2 ± √(2^2 - 4(1)(-1))] / 2

cos(x) = [-2 ± √(4 + 4)] / 2

cos(x) = [-2 ± √8] / 2

cos(x) = [-2 ± 2√2] / 2

cos(x) = -1 ± √2

So, there are two solutions for cos(x):

1. cos(x) = -1 + √2
2. cos(x) = -1 - √2

Step 7: Solve for sin(x) using the Pythagorean identity:

sin^2(x) + cos^2(x) = 1

sin^2(x) + (-1 + √2)^2 = 1 sin^2(x) + (1 - 2√2 + 2) = 1 sin^2(x) - 2√2 + 3 = 1

sin^2(x) = 2√2 - 2

sin(x) = ±√(2√2 - 2)

Therefore, the solutions for sin(x) are:

1. sin(x) = √(2√2 - 2)
2. sin(x) = -√(2√2 - 2)

So, the final solution for the equation sin^2(x) = 3cos^2(x) + sin(2x) is:

1. cos(x) = -1 + √2, sin(x) = √(2√2 - 2)
2. cos(x) = -1 + √2, sin(x) = -√(2√2 - 2)

Please note that there may be some simplifications or errors in the above steps, so it's always good to double-check the solutions.